检索结果时如何解决列名不明确的问题？

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最佳答案

You can set aliases for the columns that you are selecting:

$query = 'SELECT news.id AS newsId, user.id AS userId, [OTHER FIELDS HERE] FROM news JOIN users ON news.user = user.id'

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You can either use the numerical indices ($row[0]) or better, use AS in the MySQL:

SELECT *, user.id AS user_id FROM ...

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You can do something like

SELECT news.id as news_id, user.id as user_id ....

And then $row['news_id'] will be the news id and $row['user_id'] will be the user id

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If you don't feel like aliassing you can also just prefix the tablenames.

This way you can better automate generation of your queries. Also, it's a best-practice to not use select * (it is obviously slower than just selecting the fields you need Furthermore, only explicitly name the fields you want to have.

SELECT
news.id, news.title, news.author, news.posted,
users.id, users.name, users.registered
FROM
news
LEFT JOIN
users
ON
news.user = user.id

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I just figured this out. It's probably a bad practice but it worked for me in this case.

I am one of the lazy people who doesn't want to alias or write out every column name with a table prefix.

You can select all of the columns from a specific table by using table_name.* in your select statement.

When you have duplicated column names, mysql will overwrite from first to last. The data from the first duplicated column name will be overwritten when it encounters that column name again. So the duplicate column name that comes in last wins.

If I am joining 3 tables, each containing a duplicated column name, the order of the tables in the select statement will determine what data I am getting for the duplicate column.

Example:

SELECT table1.* , table2.* , table3.* FROM table1 LEFT JOIN table2 ON table1.dup = table2.dup LEFT JOIN table3 ON table2.dup = table3.dup;

In the example above, the value of dup I get will be from table3.

What if I want dup to be the value from table1?

Then I need to do this:

SELECT table3.* , table2.* , table1.* FROM table1 LEFT JOIN table2 ON table1.dup = table2.dup LEFT JOIN table3 ON table2.dup = table3.dup;

Now, table1 comes last, so the value of dup will be the value from table1.

I got the value I wanted for dup without having to write out every single freaking column and I still get all of the columns to work with. Yay!

I know the value of dup should be the same in all 3 tables, but what if table3 doesn't have a matching value for dup? Then dup would be blank in the first example, and that would be a bummer.

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I had this same issue with dynamic tables. (Tables that are assumed to have an id to be able to join but without any assumption for the rest of the fields.) In this case you don't know the aliases before hand.

In such cases you can first get the table column names for all dynamic tables:

$tblFields = array_keys($zendDbInstance->describeTable($tableName));

Where $zendDbInstance is an instance of Zend_Db or you can use one of the functions here to not rely on Zend php pdo: get the columns name of a table

Then for all dynamic tables you can get the aliases and use $tableName.* for the ones you don't need aliases:

$aliases = "";
foreach($tblKeys as $field)
$aliases .= $tableName . '.' . $field . ' AS ' . $tableName . '_' . $field . ',' ;
$aliases = trim($aliases, ',');

You can wrap this whole process up into one generic function and just have cleaner code or get more lazy if you wish :)

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@Jason. You are correct except that php is the culprit and not mysql. If you put your JOIN in Mysql Workbench you will get three columns with the exact same name (one for each table) but not with the same data (some will be null if that table has no match for the JOIN).

In php if you use MYSQL_NUM in mysql_fetch_array() then you will get all columns. The problem is when you use mysql_fetch_array() with MYSQL_ASSOC. Then, inside that function, php is building the return value like so:

$row['dup'] = [value from table1]

and later on...

$row['dup'] = [value from table2]

...

$row['dup'] = [value from table3]

So you will get only the value from table3. The problem is that a result set from mysql can contain columns with the same name but associative arrays in php don't allow duplicate keys in arrays. When the data is saved in associative arrays, in php, some information is silently lost...

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Another tip: if you want to have cleaner PHP code, you can create a VIEW in the database, e.g.

For example:

CREATE VIEW view_news AS
SELECT
news.id news_id,
user.id user_id,
user.name user_name,
[ OTHER FIELDS ]
FROM news, users
WHERE news.user_id = user.id;

In PHP:

$sql = "SELECT * FROM view_news";

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There are two approaches:

Using aliases; in this method you give new unique names (ALIAS) to the various columns and then use them in the PHP retrieval. e.g.

SELECT student_id AS FEES_LINK, student_class AS CLASS_LINK
FROM students_fee_tbl
LEFT JOIN student_class_tbl ON students_fee_tbl.student_id = student_class_tbl.student_id

and then fetch the results in PHP:

$query = $PDO_stmt->fetchAll();


foreach($query as $q) {
echo $q['FEES_LINK'];
}

Using place position or resultset column index; in this, the array positions are used to reference the duplicated column names. Since they appear at different positions, the index numbers that will be used is always unique. However, the index positioning numbers begins at 0. e.g.
```
SELECT student_id, student_class
FROM students_fee_tbl
LEFT JOIN student_class_tbl ON students_fee_tbl.student_id = student_class_tbl.student_id
```
and then fetch the results in PHP:
```
$query = $PDO_stmt->fetchAll();


foreach($query as $q) {
echo $q[0];
}
```

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Here's an answer to the above, that's both simple and also works with JSON results being returned. While the SQL query will automatically prefix table names to each instance of identical field names when you use SELECT *, JSON encoding of the result to send back to the webpage, ignores the values of those fields with a duplicate name and instead returns a NULL value.

Precisely what it does is include the first instance of the duplicated field name, but makes its value NULL. And the second instance of the field name (in the other table) is omitted entirely, both field name and value. But, when you test the query directly on the database (such as using Navicat), all fields are returned in the result set. It's only when you next do JSON encoding of that result, do they have NULL values and subsequent duplicate names are omitted entirely.

So, an easy way to fix that problem is to first do a SELECT *, then follow with aliased fields for the duplicates. Here's an example, where both tables have identically named site_name fields.

SELECT *, w.site_name AS wo_site_name FROM ws_work_orders w JOIN ws_inspections i WHERE w.hma_num NOT IN(SELECT hma_number FROM ws_inspections) ORDER BY CAST(w.hma_num AS UNSIGNED);

Now in the decoded JSON, you can use the field wo_site_name and it has a value. In this case, site names have special characters such as apostrophes and single quotes, hence the encoding when originally saving, and the decoding when using the result from the database.

...decHTMLifEnc(decodeURIComponent( jsArrInspections[x]["wo_site_name"]))

You must always put the * first in the SELECT statement, but after it you can include as many named and aliased columns as you want, as repeatedly selecting a column causes no problem.

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When using PDO for a database interacttion, one can use the PDO::FETCH_NAMED fetch mode that could help to resolve the issue:

$sql = "SELECT * FROM news JOIN users ON news.user = user.id";
$data = $pdo->query($sql)->fetchAll(PDO::FETCH_NAMED);
foreach ($data as $row) {
echo $row['column-name'][0]; // from the news table
echo $row['column-name'][1]; // from the users table
echo $row['other-column']; // any unique column name
}